Question Help * The table below lists weights (carats) and prices (dollars) of r

Question

Question Help * The table below lists weights (carats) and prices (dollars) of randomly selected diamonds Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval There is sufficient evidence to support a claim of a inear correlation, so it is reasonable to use the regression equation when making predictions For the prediction interval, use a 95% conhdence levei with a danond that weighs 0 8 carats 1 0 0 5 1345 0 5 $1402 Weight 0 7 ice $517 $1169 a·Find the oxplanod vanation Round to the nearest whole number as needed b. Find the unexplained varation (Round to the nearest whole number as needed ) c. Find the indicated prediction interval Round to the nearest whole number as needed )

Explanation / Answer

Let's uded minitab:

Step 1) First enter the given data set in minitab columns.

Step 2) Click on Stat>>>Regression>>>General regression...

Response: select Price

Model: select Weight

Step 3) Click on Prediction

in "New observation for continuous predictors:

we want to predict y for given Weight = 0.8

So put 0.8

Select confidence limits:

and prediction limits

Then click on OK

then click on Option.

Confidence level for all interval: put 95

Types of confidence interval: Two-sided.

then click on OK and again Click on OK

So we get the following minitab output

General Regression Analysis: Price versus Weight

Regression Equation

Price = -1991.46 + 7149.04 Weight

Coefficients

Term Coef SE Coef T P
Constant -1991.46 581.205 -3.42643 0.027
Weight 7149.04 951.221 7.51565 0.002

Summary of Model

S = 532.457 R-Sq = 93.39% R-Sq(adj) = 91.73%
PRESS = 6275611 R-Sq(pred) = 63.40%

Analysis of Variance

Source DF Seq SS Adj SS Adj MS F P
Regression 1 16014094 16014094 16014094 56.485 0.0016776
Weight 1 16014094 16014094 16014094 56.485 0.0016776
Error 4 1134042 1134042 283510
Lack-of-Fit 3 1132417 1132417 377472 232.362 0.0481783
Pure Error 1 1625 1625 1625
Total 5 17148135

Fits and Diagnostics for Unusual Observations

Predicted Values for New Observations

NewObs Fit SE Fit 95% CI 95% PI
1 3727.78 310.667 (2865.23, 4590.33)  (2016.21, 5439.35)

The explained variation is measure by coefficient of determination (R^2)

From the above output R^2 = R -sq = 93.39%

After rounding we get R^2 = 93%

So answer of part a) is 93

b) Unexplained variation = (100 - 93.39)% = 6.61% = 7%

So answer of part b) is 7

c) The 95% prediction interval for weight = 0.8 is (2016.21, 5439.35)
After rounding we get $ 2016 < y < $ 5439

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